= Solution
Start with the supplied bounded-complete, <one-consistent> <first-order theory> $T$ proving a false <first-order sentence> $F$. Replace it by its deductive closure, which preserves these properties and contains $F$. Enumerate the <arithmetic> <first-order sentences> as $X_0,X_1,\ldots$. Recursively extend the current <first-order theory> by either $X_n$ or $\neg X_n$, choosing an extension which remains <one-consistent> as permitted by the supplied lemma, and then take its deductive closure. Call the resulting increasing sequence $T_n$, with $T_0$ the initial <first-order theory>, and put $S=\bigcup_nT_n$.
Every stage is still bounded-complete. By (b), every stage has <Sigma-1 soundness>. If $S$ proved a false Sigma-1 <first-order sentence>, the finite <formal proof> would use assumptions from one stage $T_n$, contradicting its <Sigma-1 soundness>. Thus $S$ has <Sigma-1 soundness> and is bounded-complete, so (b) makes it <one-consistent>.
This <soundness of an arithmetic theory> argument is essential: the definition of <one-consistency> involves infinitely many standard refutations, so it would be insufficient merely to say that a violation has a finite <formal proof>. What has a finite <formal proof> is the false <existential sentence>, and <bounded completeness of an arithmetic theory> is what turns <one-consistency> into the needed <soundness of an arithmetic theory> condition.
The <union> is deductively closed because each <formal proof> has finitely many assumptions. It contains either $X$ or $\neg X$ for every <first-order sentence>, by the <numeralwise enumeration in arithmetic>, and contains the false <first-order sentence> $F$. If it were omega-consistent, (c) would make it correct on all <first-order sentences>, contradicting $F$. We have therefore constructed
$$
\boxed{S\text{ complete and one-consistent, but not omega-consistent}.}
$$
<Omega-consistency> always implies <one-consistency> by restricting its matrices. This counterexample proves that the converse is strictly false.
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